uniqueness of quotient rings Given some ring $R$ and two ideals $I$ and $J$ of $R$ such that $I \neq J$, is it possible for $R/I \cong R/J$? 
 A: Consider the ring $R=K[X]$ where $K$ is either $\Bbb R$, $\Bbb C$ or ${\Bbb F}_p$ (the field with $p$ elements). Ideals in $R$ are always principal.
Let $P$ and $Q$ two irreducible polynomials in $R$. Then
$$
R/(P)\simeq R/(Q) \iff \deg(P)=\deg(Q).
$$
This follows from the fact that $\Bbb C$ is the only non-trivial algebraic extension of $\Bbb R$ and the classification theorem for finite fields.
Another important example is that of $R$ the integral closure of $\Bbb Z$ in a Galois finite extension of $\Bbb Q$. In this case for any prime number $p$ we have a decomposition
$$
pR={\cal P}_1\cdots{\cal P}_t
$$
with ${\cal P}_i$ prime (and maximal) in $R$ and residue field $R/{\cal P}_i$ independent of $i$.
Obviously one can obtain many more examples, for instance from the ring of functions of algebraic varieties.
A: One ubiquitous example is provided by evaluation maps, viz. $\rm\ R[x]/(x-r)\ \cong\: R\ $ for all $\rm\ r\in R\:$
So, similar to casting out nines, we can check polynomial arithmetic $\rm\ (mod\ x)\ $ i.e. by doing the arithmetic only for the constant terms, $\:$ using $\rm\ (f+g)(0) = f(0) + g(0),\ (fg)(0) = f(0)\:g(0)\:,\: $ or we can check $\rm\: (mod\ x-1)\: $ by adding coefficients $\rm\ (f+g)(1) = f(1)+g(1),\ (fg)(1) = f(1)\:g(1)\:.\ $ In either case the arithmetic checks amount to performing computations in a (simpler) ring that is isomorphic to the coefficient ring, namely $\rm\ R[x]/(x)\ \cong\ R\ \cong\ R[x]/(x-1)\:.$
